There are only finitely many squares in the sequence because for some square big enough the offset to the next square is greater than 9^9. Now, for each number d = (largest digit of n)^(smallest digit of n) we just have to walk through the squares, check if this square has the correct (largest digit of n)^(smallest digit of n) property and added d is also a square. I have done that search computation exhaustively (using a PARI program). There are no more squares. - Maon Wenders, Jun 02 2012

This sequence has infinitely many terms because there are infinitely many numbers b^2 - 1 that contain a zero in their decimal expansion. - T. D. Noe, Jul 20 2012